$\alpha^\lambda .\beta^{(1-\lambda)}\le \lambda.\alpha+(1-\lambda).\beta\Rightarrow\sum_{i=1}^{n}|\hat x_i .\hat y_i|\le\frac{1}{p}.\sum_{i=1}^{n}|\hat x_i|^p+\frac{1}{q}.\sum_{i=1}^{n}|\hat y_i|^q$
$\hat x=\frac{x}{||x||_p}$ ; $\hat y=\frac{y}{||y||_q}$ ; $\frac{1}{p}+\frac{1}{q}=1$
I don't understand how $\sum_{i=1}^{n}|\hat x_i .\hat y_i|$ appeared.
Your question comes from part (b) of a three part question involving Holder's inequality. In part (a) you have established that for $0 < \lambda < 1$, the following inequality holds for nonnegative real numbers $\alpha$ and $\beta$,
$$ \alpha^\lambda\beta^{1-\lambda} \le \lambda\alpha + (1-\lambda)\beta. $$
In part (b) you are asked to show that with $\hat{x} = \tfrac{\vec{x}}{||\vec{x}||_p}$ and $\hat{y} = \tfrac{\vec{y}}{||\vec{y}||_q}$ and $\tfrac{1}{p} + \tfrac{1}{q} = 1$, this inequality becomes
$$ \sum\limits_{i=1}^{n}|\hat{x}_i\hat{y}_i| \le \tfrac{1}{p}\sum\limits_{i=1}^{n}|\hat{x}_i|^p + \tfrac{1}{q}\sum\limits_{i=1}^{n}|\hat{y}_i|^q. $$
If you substitute $\lambda = \tfrac{1}{p}$, $\alpha = |\hat{x}_i|^p$, and $\beta = |\hat{y}_i|^q$, then $1-\lambda = \tfrac{1}{q}$ and $\alpha^\lambda\beta^{1-\lambda} \le \lambda\alpha + (1-\lambda)\beta$ becomes
$$ (|\hat{x}_i|^p)^{1/p}(|\hat{y}_i|^q)^{1/q} \le \tfrac{1}{p}|\hat{x}_i|^p + \tfrac{1}{q}|\hat{y}_i|^q. $$
$$ |\hat{x}_i||\hat{y}_i| \le \tfrac{1}{p}|\hat{x}_i|^p + \tfrac{1}{q}|\hat{y}_i|^q. $$
Hence the following sum is obtained:
$$ \sum\limits_{i=1}^{n}|\hat{x}_i\hat{y}_i| \le \tfrac{1}{p}\sum\limits_{i=1}^{n}|\hat{x}_i|^p + \tfrac{1}{q}\sum\limits_{i=1}^{n}|\hat{y}_i|^q. $$
